When Two Samples of Water are Mixed, what Final Temperature Results?

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Problem #1:

Determine the final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8 °C.

This is problem 8a from Worksheet #2.

First some discussion, then the solution. Forgive me if the points seem obvious:

1) The colder water will warm up (heat energy "flows" into it). The warmer water will cool down (heat energy "flows" out of it).
2) The whole mixture will wind up at the SAME temperature. This is very, very important.
3) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the colder water)

This problem type becomes slightly harder if a phase change is involved. For this example, no phase change. What that means is that only the specific heat equation will be involved

Solution Key Number One: We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.

The warmer water goes down from to 46.8 to x, so this means its Δt equals 46.8 minus x. The colder water goes up in temperature, so its Δt equals x minus 14.9.

That last paragraph may be a bit confusing, so let's compare it to a number line:

To compute the absolute distance, it's the larger value minus the smaller value, so 46.8 to x is 46.8 minus x and the distance from x to 14.9 is x - 14.9.

These two distances on the number line represent our two Δt values:

a) the Δt of the warmer water is 46.8 minus x
b) the Δt of the cooler water is x minus 14.9

Solution Key Number Two: the energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

qlost = qgain


q = (mass) (Δt) (Cp)


(mass) (Δt) (Cp) = (mass) (Δt) (Cp)

With qlost on the left side and qgain on the right side.

Substituting values into the above, we then have:

(32.2) (46.8 - x)(4.184) = (32.2) (x - 14.9) (4.184)

Solve for x

Two more problems of this type

Problem #2:

A 29.5 g sample of methanol at 208.9 K is mixed with 54.3 g of methanol at 302.3 K. Calculate the final temperature of the mixture assuming no heat is lost to the containers and surroundings. The specific heat of methanol is 2.53 J g¯11

Let the final temperature be 'x.' Therefore, the Δt for the warmer methanol will be '302.3 - x' and for the colder methanol, it is 'x - 208.9.' Remember, 'x' is the final temperature and it is lower than the warmer methanol and higher than the colder methanol.


(1) (mass) (Δt) (Cp) = (mass) (Δt) (Cp)

(2) qlost on the left; qgain on the right.

Substituting and solving, we have:

(29.5) (x - 208.9) (2.53) = (54.3) (302.3 - x) (2.53)

29.5x - 6162.55 = 16414.89 - 54.3x

83.8x = 22577.44

x = 269.4 K

In case, you're not sure what happened to the 2.53, I simply divided both sides by 2.53 first.

Problem #3:

A sheet of nickel weighing 10.0 g and at a temperature of 18.0 °C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 °C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

This problem requires us to find the specific heats for nickel and iron. To do this, we will use this site. The values given are respectively, 0.54 J g¯1 °C¯1 and 0.46 J g¯1 °C¯1

Notice that the units on the site are kJ kg¯11. In addition, notice that I wrote J g¯1 °C¯1. Also, notice that there is no numerical difference when using either specific heat unit (the kJ one or the J one). In other words:

one kJ kg¯11 = one J g¯1 °C¯1

The left-hand unit is the IUPAC-approved one; the one on the right-hand side is the one in most common use.

On to the solution:

qlost = qgain


(20.0) (55.6 - x) (0.46) = (10.0) (x - 18.0) (0.54)

9.2 (55.6 - x) = 5.4 (x - 18)

511.52 - 9.2x = 5.4x - 97.2

14.6x = 608.72

x = 41.7 °C

Problem #4:

Determine the final temperature when 10.0 g of steam at 100.0 °C mixes with 500.0 grams of water at 25.0 °C.

This problem is like 9 and 10 in Worksheet #2 with one difference. The sample problem has steam and the worksheet problem have ice. The ONLY differences are the numbers used and the fact that ice is associated with "cold water" side of the equation. The techniques are the same.

The "warm water" in this case is going to do two things:

a) as a gas, condense at 100.0 °C to liquid water
b) as a liquid, the temperature goes down some unknown amount

The colder water will receive all the energy for this to happen. Keeping in mind that the heat lost by the warmer must equal the heat gained by the colder, we have this:

heat lost to cooler water by condensing steam + heat used to warm cooler water = heat gained by the cooler water

Please note again that there are TWO sources of heat energy (steam condensing, then the warm water cooling down). The total energy lost by both will equal the heat energy gained by the cooler water.

Here are the numbers:

[(40700)(10.0/18.0)] + [(10)(100 - x)(4.18)] = (500.0)(x - 25)(4.18)

Notice I used 40700 J rather than 40.7 kJ for the molar heat of vaporization.

Proceeding with the solution, I get:

2131.8x = 79041.11

and so, x = 37.1 °C.

Keep in mind that 'x' was identified with the final temperature, NOT the Δt.

Also, I did this problem with 4.18. Doing it with 4.184 gives slightly different numbers. Make sure you check with your teacher as to the values of the various constants that he/she wishes for you to use.

Problem #5:

Determine the final temperature when 18.0 g of ice at -10.0 °C mixes with 275.0 grams of water at 60.0 °C.

This is like problem 8e and several following in Worksheet #2.

The "cold water" in this case is going to do three things:

a) as a solid, warm up from -10 to zero
b) all 18 g will melt
c) as a liquid, the temperature goes up some unknown amount

The warmer water must provide all the energy for this to happen. Keeping in mind that the heat lost by the warmer must equal the heat gained by the colder (qlost = qgain), we have this:

heat to warm ice 10 degrees + heat to melt ice + heat to warm cold water by unknown amount = heat lost by the warm water

Here are the numbers:

[(18)(10)(2.06)] + [(6020)(18.0/18.0)] + [(18)(x - 0)(4.184)] = (275.0)(60.0 - x)(4.184)

Notice I used 6020 J rather than 6.02 kJ for the molar heat of fusion. That is because the other two parts of the left-hand side of the equation will give Joules as their answer. I used 6020 so that all three parts would be in Joules. If I had used 6.02, then that middle part woud have been in units of kJ.

Notice the use of x - 0 and 60.0 - x. This time, visualize (or write out) the number line used above. The zero is to the left, the 60.0 to the right and the x is in between the 0 and the 60.0.

Proceeding with the solution, I get:

1225.912x = 62945.2

and so, x = 51.3 °C.

Keep in mind that 'x' was identified with the final temperature, NOT the Δt.

Problem #6:

55.0 mL of ethanol (d = 0.789 g/mL) at 8.0 °C is mixed with 55.0 mL of water at 28.2 °C. Assuming no heat is lost, what is the final temperature of the mixture?

Two things must be done first: (1) determine how many grams of ethanol are present and (2) determine the specific heat for ethanol.

55.0 mL times the density (0.789 g/mL) gives the mass of ethanol present. This value is 43.395 g. I will use this value and round off at the end of the calculation.

The specific heat value needs to be looked up on the Internet. The most common value found in a search is 2.44 J g¯1 °C¯1, so that is what we will use.

Letting 'x' equal the final temperature, we have this:

(43.395) (x - 8.0) (2.44) = (55.0) (28.2 - x) (4.184)

You may finish this problem.

Problem #7: 900.0 L of palm oil at 47.0 °C is mixed with 200.0 L of palm oil at 76.0 °C. What is the final temperature? Assume no heat is lost to the surroundings.

Notice that no specific heat for palm oil is provided. That is because none is needed. Also, a bit of Internet sleuthing shows that the density of palm oil is about 0.9 g/cm3.

1) Let us convert 900.0 L and 200.0 L to grams using the density:

900,000 mL x 0.9 g/cm3 = 8.10 x 105 g

200,000 mL x 0.9 g/cm3 = 1.80 x 105 g

Remember that 1 mL equals 1 cm3.

2) We will let 'x' equal the final temperature, so the two temperature changes are:

the 900 L goes up in temperature, so use 'x - 47.0'

the 200 L goes down in temperature, so use '76.0 - x'

3) Keeping in mind that the heat lost by the 200 L equals the heat gained by the 900 L, we write:

(1.80 x 105) (76.0 - x) = (8.10 x 105) (x - 47.0)

x = 52.3 °C

I did not write the specific heat values because they would just cancel out. This is because we have palm oil on both sides of the equation. If the substances were different (see examples above), the two specific heats would have to be included.

Also, when I did this problem on the calculator, I dropped the 105 portion of the mass. In addition, if you were to ignore the use of the density and use, say 200 and 900, you'd get 51.0 °C for the answer.

Problem #8: A 2.00 x 102 g brass block at 85.0 °C is placed in a plastic foam cup containing 2.00 x 102 g of water at 50.0 °C. No heat is lost to the cup or the surroundings. Find the final temperature of the mixture.

The solution to this problem requires the specific heat of brass. Use this site to find the value. See my comments (in problem #3 above) on the specific heat unit used.

The solution is left to the reader.

Problem #9: 364 g of water at 34.0 °C is added to ice at 0.0 °C. If the final temperature of the system (which you can assume is isolated) is 0.0 °C, determine how much ice melted. The specific heat of water is 4186 J/kg ⋅ °C. The latent heat of fusion for water is 335,000 J/kg.


1) Determine energy lost by warm water:

q = (mass) (Δt) (Cp)

q = (0.364 kg) (34.0 °C) (4186 J/kg ⋅ °C)

q = 51805.936 J

2) Determine how much ice is melted by 51805.936 J:

51805.936 J divided by 335,000 J/kg = 0.155 kg = 155 g (to three sig figs)

Problem #10: How much ice (in grams) would have to melt to lower the temperature of 353.0 mL of water from 26.0 °C to 6.0 °C?


1) Determine heat lost by cooling water from 26.0 to 6.0:

q = (353.0 g) (20.0 °C) (4.184 J g¯1 °C¯1)

q = 29539.04 J

Note the silent conversion from volume of water to mass of water using the density of 1.00 g/mL.

2) The heat lost by the warm water does two things:

a) melt an unknown mass of ice
b) take the same mass of melted ice from zero Celsius to 6 °C

3) We can express it thusly:

29539.04 J = (6020 J/mol) (x/18.015 g/mol) + (x) (6.0 °C) (4.184 J g¯1 °C¯1)

29539.04 J = 334.166x + 25.104x

x = 82.2 g

Problem #11: An unknown volume of water at 18.2 °C is added to 27.8 mL of water at 33.6 °C. If the final temperature is 23.5 °C, what was the unknown volume? (Assume that no heat is lost to the surroundings; d of water is 1.00 g/mL.)


Note the our 'x' in this problem will be an unknown mass of water. Since we know the starting temperatures and the final temperature, we can calculate the Δt values.

1) Calculate the two Δt values:

cooler water ⇒ 23.5 minus 18.2 = 5.2
warmer water ⇒ 33.6 minus 23.5 = 10.1

2) qlost = qgain:

(x) (5.3) (4.184) = (27.8) (10.1) (4.184)

5.3x = 280.78

x = 53.0 g

This is 53.0 mL of water.

Problem #12: For two identical blocks, Tf is the average of the initial temperatures, so that Tf = 1/2 (T1 + T2). Show, for a system of two blocks totally isolated from the surroundings, that this is true. (Hint: since the blocks are of the same material, they will have the same Cp.)


I will use T1 for the initial temperature of the warm block and T2 for the initial temperature of the cold block.

heat lost by warm block = heat gained by cold block

(mass) (ΔTwarm) (Cp) = (mass) (ΔTcold) (Cp)

for identical blocks, mass = mass and Cp = Cp; therefore:

ΔTwarm = ΔTcold

T1 - Tf = Tf - T2

2Tf = T1 + T2

Tf = 1/2 (T1 + T2)

Problem #13: A student mixed 6.00 g of ice at -3.4 °C with 1.00 g of steam at 103.0 °C. What is the final temperature of this mixture?


heat gained in warming up = heat lost in cooling down

heat gained by ice + heat to melt ice + heat to raise water temp = heat lost by steam + heat lost as steam condenses + heat lost as water cools

(6.00 g) (3.4 °C) (2.06 J/g °C) + (6.00 g / 18.0 g/mol) (6020 J/mol) + (6.00 g) (x - 0 °C) (4.184 J/g °C) = (1.00 g) (3.0 °C) (2.02 J/g °C) + (1.00 g / 18.0 g/mol) (40700 J/mol) + (1.00 g) (100 - x °C) (4.184 J/g °C)

42.024 + 2006.667 + 25.104x = 6.06 + 2261.11 + 418.4 - 4.184x

29.288x = 636.879

x = 21.7 °C

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Worksheet #2